Off-by-one in general term
Very CommonFORMULA
The general term formula gives Tr+1, not Tr. The 5th term uses r=4.
Why: The binomial expansion is indexed from r=0, so Tr+1=nCr⋅an−r⋅br.
WRONG: 5th term of (a+b)n: T5=nC5⋅an−5⋅b5 RIGHT: 5th term: T5=T4+1=nC4⋅an−4⋅b4. Always use r=term number−1. See pattern: Coefficient / Term Finding →Missing negative signs in expansion
Very CommonSIGN ERROR
When expanding (a−b)n, the sign alternates. Tr+1 has factor (−1)r.
Why: The negative sign on b gets raised to the power r, creating alternating signs.
WRONG: T3 of (x−2)5=5C2⋅x3⋅4=40x3 RIGHT: T3=5C2⋅x3⋅(−2)2=40x3. Here it is positive, but for odd r it is negative. Track (−1)r carefully. Integral terms: both exponents must be integers
CommonCASE MISS
For a term to be rational, BOTH exponents in the general term must be non-negative integers.
Why: Students check only one exponent and miss the constraint on the other.
WRONG: In (21/3+31/4)12, checking only that r is a multiple of 4 RIGHT: Need r to be a multiple of 4 AND (12−r) to be a multiple of 3. Check both conditions simultaneously. See pattern: Sum of Coefficients →Sum of coefficients vs binomial coefficients
CommonFORMULA
Sum of binomial coefficients of (1+x)n is 2n. Sum of coefficients of f(x) is f(1).
Why: Students apply the 2n shortcut to all expansions, not just (1+x)n.
WRONG: Sum of coefficients of (1−3x+10x2)n=2n RIGHT: Put x=1: (1−3+10)n=8n. The sum of coefficients equals f(1), not always 2n. See pattern: Properties of Binomial Coefficients →Remainder: wrong base decomposition
CommonSIGN ERROR
When finding remainder of anmodm, the decomposition must use the correct sign.
Why: Incorrect splitting of the base leads to wrong binomial expansion and wrong remainder.
WRONG: 64kmod9: writing 64=9×7+1, then remainder =1k (correct here, but sign errors are common) RIGHT: Always verify: base = divisor × quotient + remainder. 64=9×7+1 ✓. Then expand (9×7+1)k. See pattern: Remainder & Divisibility →